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We cannot plug in zero and 1 as the expression (x+1) (|x| - 1) will not hold true. All numbers greater than 1 will hold true for the expression. All decimals and negative numbers will make the expression negative.

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If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Thanks Bunnel.I have a question regarding the modulus of X. |X| = -X when X<= 0|X| = X when X> 0Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.So, is |X| = -X when X< 0|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)

So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.Hope the description of my question is clear enough!Thank you in advance!

My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)

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If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Thanks Bunnel.I have a question regarding the modulus of X. |X| = -X when X<= 0|X| = X when X> 0Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.So, is |X| = -X when X< 0|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)

So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.Hope the description of my question is clear enough!Thank you in advance!

My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)

You can include 0 in either of the ranges:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\).

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

The point is that |0|=0, so it doesn't matter in which range you include it.

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Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 02) x-1 > 03) -x-1 > 0

If correct, could you please demonstrate on how to proceed?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
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This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.

Fair enough Bunuel, but just for the sake of understanding and learning the concept.

If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.

Fair enough Bunuel, but just for the sake of understanding and learning the concept.

If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?

Kindly pardon me. I am unable to understand your explanation. Could you please explain how you arrived at the 4 equations? I think the 4 cases have come up because x could be either positive or negative. Is it so?

Further you've mentioned that we've to consider 2x2 cases. I understand that you have taken 3 cases where x < 0 and 1 case where x > 0. I'm totally lost now.

Kindly pardon me. I am unable to understand your explanation. Could you please explain how you arrived at the 4 equations? I think the 4 cases have come up because x could be either positive or negative. Is it so?

Further you've mentioned that we've to consider 2x2 cases. I understand that you have taken 3 cases where x < 0 and 1 case where x > 0. I'm totally lost now.

Please help!

I had typos there. Edited now. Anyway the point is that we consider two cases for |x| and then two cases for the multiples both negative and both positive.
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If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.

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If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.

\(-(x+1)^2>0\);

Add (x+1)^2 to both sides: \(0>(x+1)^2\), which is the same as \((x+1)^2<0\).
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06 Dec 2014, 13:52

1

This post receivedKUDOS

Is x > 1?

I would approach it the following way :

(1) (x+1) (|x| - 1) > 0

(+) * (+) > 0 or (-) * (-) > 0

For both of the parts to be positive we can see that x >1 . Just by trying few values you can figure this out. X cant be Zero as then the second part becomes - . X cant be 1 as then second part becomes 0 and hence the whole LHS becomes Zero.

For both of the parts to be negative we try any value less an Zero and see that no value will satisfy the equation. Hence X cannot be negative.. Hence A is Sufficient.

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(x+1)(|x|-1)=0 has x=1 and x=-1 as its solution. However, x=-1 is a repeated root. Hence, on the number line, the sign will not change when it passes x=-1.

--------(-1)---------(1)+++++++

Therefore, only when x>1, (x+1)(|x|-1)>0.

Is my solution correct? I remember my highschool math teacher said something like 'the sign doesn't change when it passes a double root'. But, it's years ago and I just want to make sure that the approach is valid.